In Hilbert space $u$, Let $T_1$,$T_2$ is selfadjoint operator, if exit $c>0$ such that $cI\le T_1\le T_2$, prove $T_1$,$T_2$ have a bounded inverse operator and $c^{-1}I\ge T_1^{-1}\ge T_2^{-1}$.
I have proved $T_1$ and $T_2$ have inverse operators by using the relation between the range of the adjoint operator and the null space of the operator. How do I prove the second half of the problem and the boundedness of the inverse operator?
$\newcommand\ip[2]{\langle#1,#2\rangle}$If $T$ is an self-adjoint operator such that $T\ge cI$ for some real $c>0$, then, as you showed, $T$ is invertible. So, for any $x$ in your Hilbert space and $y:=T^{-1}x$, $$\|x\|^2=\ip xx=\ip{Ty}{Ty}=\ip{T^2y}{y}\ge c^2\ip yy=c^2\|y\|^2 =c^2\|T^{-1}x\|^2,$$ so that $\|T^{-1}\|\le c^{-1}<\infty$.
Let now $T_{1+s}:=T_1+s(T_2-T_1)$. Everywhere here, $s\in[0,1]$. Then $T_{1+s}^{-1}$ exists and $\|T_{1+s}^{-1}\|\le c^{-1}$, because $T_{1+s}$ is an self-adjoint operator and $T_{1+s}\ge cI$.
Now take any $x$ in your Hilbert space and let $$g_x(s):=\ip{T_{1+s}^{-1}x}{x}.$$ Noting that $T^{-1}-S^{-1}=T^{-1}(S-T)S^{-1}$ for any invertible operators $T$ and $S$, we get $$\frac d{ds}g_x(s)=\ip{T_{1+s}^{-1}(T_1-T_2)T_{1+s}^{-1}x}{x}\le0,$$ since $T_1-T_2\le0$. So, $g_x(s)$ is nonincreasing in $s$. So, for all $x$, $$\ip{T_1^{-1}x}{x}=g_x(0)\ge g_x(1)=\ip{T_2^{-1}x}{x}.$$ So, $T_1^{-1}\ge T_2^{-1}$. $\quad\Box$
